Exercise 11.3
Question 7
यदि (If) $y=x+\frac{1}{x}_{+\frac{1}{x}_{+\frac{1}{x}_+\dots \text{ to } \infty}}$ सिद्ध करें कि( prove that) $\left(x^{2}-y^{2}+3\right) \frac{d y}{d x}=1$Sol :
$y=x+\frac{1}{x}_{+\frac{1}{x}_{+\frac{1}{x}_+\dots \text{ to } \infty}}$
$y=x+\frac{1}{y}$
Differentiating with respect to x
$\frac{d y}{d x}=1-\frac{1}{y^{2}} \cdot \frac{d y}{d x}$
$\frac{d y}{dx}+\frac{1}{y^{2}} \frac{d y}{d x}=1$
$\left[1+\frac{1}{y^{2}}\right] \frac{d y}{d x}=1$
$\left[y^{2}-2 \cdot y \cdot \frac{1}{y}+\frac{1}{y^{2}}-y^{2}+3\right] \frac{d y}{d x}=1$
$\left[\left(y-\frac{1}{y}\right)^{2}-y^{2}+3\right] \frac{d y}{d x}=1$
$\left[x^{2}-y^{2}+3\right] \frac{d y}{d x}=1$
Question 8
यदि (If) $y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots . \text { to } \infty}}}$ सिद्ध करे कि (prove that) $\frac{d y}{d x}=\frac{1}{2 y-1}$Sol :
$y=\sqrt{x+\sqrt{x+\sqrt{x+\ldots . \text { to } \infty}}}$
$y=\sqrt{x+y}$
Squaring both sides
$y^{2}=x+y$
Differentiating with respect to x
$2 y.\frac{d y}{dx}=1+\frac{d y}{d x}$
$2 y \frac{d y}{d x}-\frac{d y}{d x}=1$
$(2 y-1) \frac{d y}{d x}=1$
$\frac{d y}{d x}=\frac{1}{2 y-1}$
Question 9
यदि (If) $\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$ सिद्ध करें कि (prove that) $\frac{d y}{d x}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$Sol :
$\sqrt{1-x^{2}}+\sqrt{1-y^{2}}=a(x-y)$
Differentiating with respect to x
$\frac{1}{2 \sqrt{1-x^{2}}} \times(-2 x)+\frac{1}{2 \sqrt{1-y^{2}}} \times(-2 y) \cdot \frac{d y}{d x}=a\left[1-\frac{dy}{d x}\right]$
$\frac{-x}{\sqrt{1-x^{2}}}-\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}=a-\frac{a d y}{d x}$
$a \frac{d y}{d x}-\frac{y}{\sqrt{1-y^{2}}} \frac{d y}{d x}=a+\frac{x}{\sqrt{1-x^{2}}}$
$\left[a-\frac{y}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=a+\frac{x}{\sqrt{1-x^{2}}}$
$\left[\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}-\frac{y}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{\sqrt{1-x^{2}}+\sqrt{1-y^{2}}}{x-y}+\frac{x}{\sqrt{1-x^{2}}}$
$\left[\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}+1-y^{2}-x y+y^{2}}{(x-y) \sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{1-x^{2}+\sqrt{1-x^{2}} \sqrt{1-y^{2}}+x^{2}-xy}{(x-y) \sqrt{1-x^{2}}}$
$\left[\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}-x y+1}{\sqrt{1-y^{2}}}\right] \frac{d y}{d x}=\frac{\sqrt{1-x^{2}} \sqrt{1-y^{2}}-x y+1}{\sqrt{1-x^{2}}}$
$\frac{dy}{d x}=\frac{\sqrt{1-y^{2}}}{\sqrt{1-x^{2}}}=\sqrt{\frac{1-y^{2}}{1-x^{2}}}$
Question 10
यदि (If) $y=x+\frac{1}{x}$ सिद्ध करें कि (prove that) $x \frac{d y}{d x}+y=2 x$ $x \frac{d y}{d x}+y=2 x$Sol :
$y=x+\frac{1}{x}$
$y=\frac{x^{2}+1}{x}$
$x y=x^{2}+1$
Differentiating with respect to x
$1 \cdot y+x \frac{d y}{d x}=2 x+0$
$x \frac{d y}{d x}+y=2 x$
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